A completely positive equivariant map $\varphi: A \to B$ induces a map $A \rtimes_r \Gamma \to B \rtimes_r \Gamma.$

Recall the construction of the reduced crossed product:

Let $$\Gamma$$ be a discrete group and $$A$$ be a $$C^*$$-algebra with an action $$\alpha: \Gamma\to \operatorname{Aut}(A)$$. Consider the $$*$$-algebra $$C_c(\Gamma,A)$$ of finitely supported functions $$\Gamma \to A$$ with the $$\alpha$$-twisted multiplication and involution. Then we can build a canonical faithful representation of $$C_c(\Gamma,A)$$ as follows: start with a faithful representation $$A \subseteq B(H)$$. This induces a new faithful representation $$\pi: A \to B(H \otimes \ell^2(\Gamma))$$ by $$\pi(a)(\xi \otimes \delta_g) = \alpha_{g}^{-1}(a)\xi \otimes \delta_g$$. Considering the left regular representation $$\lambda: \Gamma \to U(\ell^2(H)): g \mapsto (\delta_h \mapsto \delta_{gh})$$, we obtain an induced faithful representation $$C_c(\Gamma,A) \to B(H \otimes \ell^2(\Gamma)): \sum_{s \in \Gamma} a_s s \mapsto \sum_{s \in \Gamma} \pi(a_s)(1 \otimes \lambda_s)$$ which induces a $$C^*$$-norm on $$C_c(\Gamma,A)$$. The reduced crossed product $$A \rtimes_r \Gamma$$ is the $$C^*$$-completion of $$C_c(\Gamma,A)$$ with respect to this norm, and does not depend on the choice of faithful representation $$A\subseteq B(H)$$.

Let $$\Gamma$$ be a discrete group and let $$\varphi: A \to B$$ be a $$\Gamma$$-equivariant completely positive contraction between the $$\Gamma$$-$$C^*$$-algebras $$A$$ and $$B$$. I want to show the following:

The induced map $$C_c(\Gamma,A) \to C_c(\Gamma,B): \sum_{s \in \Gamma} a_s s \mapsto \sum_{s \in \Gamma}\varphi(a_s)s$$ is bounded, hence extends uniquely to a map $$\varphi \rtimes_r \Gamma: A \rtimes_r \Gamma \to B \rtimes_r \Gamma.$$

Attempt: Let $$\pi_A: A \to B(H_A \otimes \ell^2(\Gamma))$$ and $$\pi_B: B \to B(H_B \otimes \ell^2(\Gamma))$$ be faithful representations as above. Then by the $$C^*$$-identity. $$\|\sum_s \varphi(a_s)s\|^2 = \|\sum_s \pi_B(\varphi(a_s))(1 \otimes \lambda_s)\|^2$$ $$=\|\sum_{s,t} (1 \otimes \lambda_{s^{-1}}) \pi_B(\varphi(a_s^*)\varphi(a_t)) (1\otimes \lambda_t)\|.$$

This looks like something we could apply Cauchy-Schwarz for completely positive maps on, but the surrounding factors $$1 \otimes \lambda_{s^{-1}}$$ and $$1 \otimes \lambda_t$$ complicate this. Maybe I need to apply Cauchy-Schwarz on some carefully crafted matrix. Does anybody see how I can continue?

Of course, other approaches are also welcome! I am also interested in the following: if $$\varphi$$ is injective, then is the extension $$A \rtimes_r \Gamma \to B \rtimes_r \Gamma$$ also injective?

Well, it might perhaps help to look at this section of the book. Indeed, Proposition 4.1.5, and its proof, shows that if $$F\subseteq\Gamma$$ is a finite-set, and $$P:\ell^2(\Gamma)\rightarrow \ell^2(F)$$ the projection, then $$(1\otimes P)x(1\otimes P) = \sum_{s\in\Gamma} \sum_{p\in F\cap sF} \alpha_{p^{-1}}(a_s)\otimes e_{p, s^{-1}p} \in A\otimes M_F(\mathbb C) \cong M_F(A).$$ Here $$x = \sum a_s\lambda_s \in C_c(\Gamma, A) \subseteq A \rtimes_r \Gamma \subseteq \mathcal B(H\otimes\ell^2(\Gamma))$$ and $$e_{p,q}$$ are the matrix units of $$M_F(\mathbb C)$$. The point of this is to show that the norm of $$x$$, acting on $$H\otimes\ell^2(\Gamma)$$, does not depend upon the particular representation $$A\subseteq\mathcal B(H)$$ (because $$M_F(A)$$ has a unique norm).
However, this formula also gives us a convenient way to compute the norm of $$\varphi\rtimes 1$$. For, if $$y=(\varphi\rtimes 1)x$$, then $$y=\sum_s \varphi(a_s)\lambda_s$$, and so $$(1\otimes P)y(1\otimes P) = \sum_{s\in\Gamma} \sum_{p\in F\cap sF} \beta_{p^{-1}}(\varphi(a_s))\otimes e_{p, s^{-1}p} = \sum_{s\in\Gamma} \sum_{p\in F\cap sF} \varphi(\alpha_{p^{-1}}(a_s))\otimes e_{p, s^{-1}p},$$ the first equality by applying to above to $$B\rtimes_r\Gamma$$, and the second equality using that $$\varphi$$ is equivariant. However, this then equals $$(\varphi\otimes 1_F)\big( (1\otimes P)x(1\otimes P) \big),$$ where $$(\varphi\otimes 1_F)$$ is the dilated map $$M_F(A)\rightarrow M_F(B)$$. As $$\varphi$$ is CCP by assumption, $$\varphi\otimes 1_F$$ is contractive, and so $$\| (1\otimes P)y (1\otimes P) \| \leq \|(1\otimes P)x(1\otimes P)\| \leq \|x\|.$$ Taking the SOT limit as $$F$$ increases, so $$P\rightarrow 1$$, shows that $$\|y\| \leq \|x\|$$ as required.